Problem 40 - Calculus Early Transcendentals

Question

Sketch each region and use a double integral to find its area. The region bounded by the cardioid \(r=2(1-\sin \theta)\)

Step-by-Step Solution

Answer

Answer: The area of the region bounded by the cardioid \(r = 2(1-\sin\theta)\) is \(8\pi\).

Step 1: Sketch the cardioid region

To begin, let's first sketch the given cardioid \(r=2(1-\sin\theta)\). Recognize that it's a cardioid shape due to the form of the equation. When plotting a cardioid, we'll start by examining its properties. - When \(\theta = 0\), \(r = 2(1 - \sin(0)) = 2\). The cardioid begins at the point \((2,0)\). - When \(\theta = \frac{\pi}{2}\), \(r = 0\). The cardioid intersects the origin. - When \(\theta = \pi\), \(r = 2(1-\sin(\pi)) = 2\). The cardioid has a maximum distance from the origin at \(\theta = \pi\). - When \(\theta = \frac{3\pi}{2}\), \(r = 0\). The cardioid intersects the origin again. - When \(\theta = 2\pi\), \(r = 2(1-\sin(2\pi)) = 2\). We have made a full rotation and arrived back at the starting point. With these key points, we can sketch the cardioid region. The region appears somewhat like a teardrop nearby the positive x-axis.

Step 2: Convert the polar equation to Cartesian form

To find the area of the cardioid using a double integral, we must first convert the polar equation \(r=2(1-\sin\theta)\) to its Cartesian form. We will use the conversion formulas \(x = r\cos\theta\) and \(y = r\sin\theta\). From \(r=2(1-\sin\theta)\), we get \(r = 2 - 2\sin\theta\). Now, multiply both sides of this equation by \(r\): \(r^2 = r(2 - 2\sin\theta)\) Substitute \(x^2 + y^2\) for \(r^2\), \(x\) for \(r\cos\theta\), and \(y\) for \(r\sin\theta\): \(x^2 + y^2 = x(2 - 2y)\) Now, we have the polar equation in Cartesian coordinates.

Step 3: Set up the double integral

To find the area of the region, we will integrate \(r = 2 - 2\sin\theta\) with respect to \(r\) and \(\theta\). The limits for the \(r\) integral will be from 0 to \(2(1-\sin\theta)\), and the limits for the \(\theta\) integral will be from 0 to \(2\pi\). The set up will be: \(\int_0^{2\pi}\int_0^{2(1-\sin\theta)} r \, dr \, d\theta\)

Step 4: Evaluate the double integral

Proceed by evaluating the inner integral first: \(\int_0^{2\pi}\left[\frac{1}{2}r^2\right]_0^{2(1-\sin\theta)}\,d\theta = \int_0^{2\pi} (2 - 2\sin\theta)^2 \, d\theta\) Now, evaluate the outer integral. Expand the square, integrate, and find the value: \(\int_0^{2\pi}(4 - 8\sin\theta + 4\sin^2\theta)\,d\theta = \left[4\theta - 8\cos\theta + \frac{4}{3}\sin(2\theta)\right]_0^{2\pi}\) Evaluating this at the limits and subtracting, we get: \(4(2\pi) - 0 + 0 - 0 = 8\pi\) Therefore, the area of the region bounded by the cardioid \(r=2(1-\sin\theta)\) is \(8\pi\).

Key Concepts

Polar Coordinates

Polar coordinates offer a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction. Unlike the rectangular Cartesian coordinate system, which uses the x and y coordinates, polar coordinates use the radius (r) and angle (θ), often measured in radians.

A polar coordinate system is especially useful in situations where relationships are more naturally expressed in terms of distances and angles. They are commonly used in navigation, engineering, and physics. To visualize a polar coordinate, imagine the radius as a line extending from the origin (the reference point) to the point in question, and the angle as the measure from the positive x-axis to the radius line, counterclockwise.

Cardioid Equation

A cardioid is a type of curve that resembles the shape of a heart or an apple. It is a special case of an epicycloid and can be defined in polar coordinates. The general equation of a cardioid can be written as \(r = a(1 - \text{cos}(\theta))\) or \(r = a(1 - \text{sin}(\theta))\), where 'a' is a constant that determines the size of the cardioid. The cardioid has interesting properties such as having a cusp at the origin and being symmetrical about the horizontal axis if defined by cosine, or vertical axis if defined by sine, indicative of its mirrored curves along a central axis.

In our exercise, the cardioid equation is \(r = 2(1 - \text{sin}(\theta))\). By plotting different values of θ between 0 and \(2\text{π}\), we trace out the cardioid shape. Notably, the equation gives us important points which help in sketching the overall figure, essential before evaluating the area under the curve, and incidentally also helps in setting up the limits for integration when using polar coordinates.

Converting Polar to Cartesian Coordinates

To convert equations from polar to Cartesian coordinates or vice versa, standard conversion formulas are used. The x-coordinate is found by multiplying the radius by the cosine of the angle, expressed as \(x = r \text{cos}(\theta)\), and the y-coordinate is found by multiplying the radius by the sine of the angle, or \(y = r \text{sin}(\theta)\).

In the case of a cardioid or any polar graph, converting to Cartesian form can be necessary for certain types of calculations, especially if the double integral is more easily evaluated in Cartesian coordinates. By squaring both sides of the polar equation and then substituting \(x\) and \(y\) using the aforementioned conversion formulas, the equation is transformed into a form that's compatible with x-y Cartesian coordinates. It's important to understand these conversions, as the ability to switch between coordinate systems is a valuable tool in mathematics and its applications.

Evaluating Double Integrals

Double integrals extend the concept of an integral to functions of two variables. They are useful for calculating areas, volumes, and other quantities that require integration over a two-dimensional area. When faced with evaluating double integrals, it's key to correctly set up the limits of integration and the order of integration according to the region of interest.

In polar coordinates, the differential element of area is not merely \(dx\) times \(dy\) as in Cartesian coordinates, but \(r dr d\theta\). This is because the arc length element changes with the radius. The double integral, therefore, becomes \(\text{∫}\text{∫} f(r, \theta) r dr d\theta\). Solving double integrals usually involves finding the antiderivative with respect to 'r' first (the inner integral), followed by the antiderivative with respect to 'θ' (the outer integral). Successful evaluation results in the computation of area, as we've seen in our original exercise, which is clear cut when proper limits and correct formulation of the function are applied.